Paper published in POAC 99, Proc. of the 15th International Conference on Port and Ocean
Engineering under Arctic Conditions, Helsinki University of Technology, Finland, August 1999.
ENERGY BASED ICE COLLISION FORCES
Claude Daley1
1
Memorial University of Newfoundland, St. John’s, Newfoundland, Canada A1C 3X5
ABSTRACT
Ice collision forces can be determined by energy considerations. A variety of interaction
geometry cases are considered. The indentation energy functions for eight different cases are
derived and expressed in a common format. The indentation functions are expressed as
functions of the indentation model parameters, assuming a pressure-area representation. Two
types of collisions are identified; simple impacts which can be treated as equivalent to onedimensional collisions, and beaching collisions which involve two-dimensional behaviour
(indentation and sliding). Solutions for the impact cases are presented for all geometry cases.
A solutions procedure is presented for the beaching collision, with an exact solution for a
linear case. Design equation development and future directions are discussed.
1. INTRODUCTION
Ice forces on ships and structures are typically the result of collisions. The magnitude of the
force is determined by some form of limit (see e.g. Croasdale, 1980, Daley, Tuhkuri and Riska
1998). In some cases the ice strength is the determining factor, while in others the force may
be limited by available kinetic energy. In such cases the available kinetic energy is expended
in crushing (irrecoverable) and potential (recoverable) energy. Energy methods provide a
simple method of determining forces, and have long been used to do so (see Popov et. al.
1967). This paper will summarize the general energy approach, derive some old and new
cases and provide examples.
2. GENERAL APPROACH
The problem under discussion is one of impact between two objects. It is assumed that one
body is initially moving (the impacting body) and the other is at rest (the impacted body). This
concept applies to a ship striking an ice edge, or ice striking an offshore structure. The energy
approach is based on equating the available kinetic energy with the energy expended in
crushing and potential energy:
KE e = IE + PE
(1)
The available kinetic energy is the difference between the initial kinetic energy of the
impacting body and the total kinetic energy of both bodies at the point of maximum force. If
the impacted body has finite mass it will gain kinetic energy. Only in the case of a direct
(normal) collision involving one infinite (or very large) mass will the effective kinetic energy
be the same as the total kinetic energy. In such a case all motion will cease at the time of
maximum force. The indentation energy is the integral of the indentation force Fn on the
crushing indentation displacement ζc ;
ζm
IE = ∫ Fn ⋅ dζ c
0
(2)
The potential energy is the energy that has been expended in recoverable processes, which can
be either rigid body motions (pitch/heave) or elastic deformation (of either body). The
potential energy is the integral of the indentation force Fn on the recoverable displacement ζe :
ζ
PE = ∫ Fn ⋅ dζ e
(3)
0
These equations are the basis of all solutions. Equation (1) can be solved for Fn provided that
the required kinematic and geometric values are known. The general approach to determining
IE and PE will be described next, with specific geometric examples further on. After that the
determination of collision forces will be discussed.
3. ICE INDENTATION
In order to pose and solve the general energy equations it is necessary to formulate an
equation relating force to indentation. By using the pressure-area relationship to describe the
ice pressures, it is easy to derive a force-indentation relationship. This assumption means that
ice force will depend only on indentation. In this case the maximum force occurs at the time
of maximum penetration. The collision geometry is the ice/structure overlap geometry. The
average pressure Pav in the nominal contact area A is related to the nominal contact area as;
Pav = P0 ⋅ A ex
(4)
where Po is the pressure at 1m2, and ex is a constant.
The ice force is also related to the nominal contact area;
Fi = Pav ⋅ A = P0 ⋅ A1+ ex
(5)
The available kinetic energy may be the total kinetic energy, in the case of a head-on
collision, in which all motion ceases at the point of maximum force. Alternatively the
available energy may be the ‘normal’ or ‘effective’ kinetic energy, as in the case of a glancing
collision.
4. INDENTATION ENERGY
For each geometric case, there is a relationship between the normal indentation ζn and normal
contact area;
An = f A (ζ n )
(6)
where fA is a function that depends on the contact geometry. This results in a function
relating force to indentation;
Fn = f F (ζ n )
(7)
where fF = Po ( fA(ζn) )1+ex . The next step is to determine the indentation energy IE, which is
found by integrating the force;
IE = ∫ Fn dζ n = f IE (ζ n )
(8)
where fIE is a function giving the indentation energy.
5. POTENTIAL ENERGY
For each geometric/kinematic case, there may be a relationship between the normal force Fn
and potential energy. In the case of ramming the vertical component of the indentation force
results in potential energy in pitch/heave. This can be expressed in terms of indentation as;
PE = f PE (ζ n )
(9)
where fPE is a function giving the indentation energy.
6. INDENTATION GEOMETRY CASES
The relationship between indentation and nominal crushing area depend on the collision
geometry. The following cases apply to both ship-ice and ice-structure collision problems.
The first case will be derived in detail. Other cases are summarized for the sake of brevity.
6.1 Case 1 : Symmetric V Wedge
Figure 1 shows a symmetric wedge-shaped indentation in a square edge. The indentation
energy is derived as follows. The projected areas, vertical, horizontal and normal are;
tan(α )
cos 2 (γ )
tan(α )
Ah = ζ n2
2
cos (γ ) tan(γ )
Av = ζ n2
(10)
(11)
An = ζ n2
tan(α )
cos (γ ) sin(γ )
(12)
2
The normal force is related to the normal area by the pressure/area relation. The average
pressure and force are expressed as in (4) and (5). Substituting (12) into (5) we arrive at:
⎛
⎞
tan(α )
⎟⎟
Fn = p o ⎜⎜
2
⎝ cos (γ ) sin(γ ) ⎠
1+ ex
⋅ ζ n2 + 2 ex
(13)
The indentation energy is found by substituting (13) into (8), to give:
⎞
po ⎛
tan(α )
⎜⎜
⎟
IE =
2
3 + 2 ⋅ ex ⎝ cos (γ ) sin(γ ) ⎟⎠
1+ ex
ζ n3+ 2 ex
(14)
6.2 Other Cases
Figure 2 shows a symmetric spoon-shaped indentation in a square edge. Figure 3 shows a
right-angle wedge indentation. Figure 4 shows a general wedge-shaped edge indentation
(normal to hull). Figure 5 shows a general round ice edge indentation. Figure 6 shows a
general round cylinder indentation. Figure 7 shows a general rectangular cylinder indentation.
Figure 8 shows a spherical indentation.
.
where
Figure 1. Symmetric V Wedge Indentation
c=Beam/(2 LBbex)
Beam is the beam of the vessel
LB is the bow length
Figure 2. Symmetric Spoon Indentation
Figure 3. Right-apex Oblique Indentation
Figure 4. General Wedge-shaped Edge
(normal to hull).
Figure 5. General Round Edge .
Figure 6. Round Vertical Cylinder
Figure 7. Rectangular Vertical Cylinder
Figure 8. Spherical Contact
7. SUMMARY OF CASES
In each case the force and indentation energy can be stated as;
Fn = p o ⋅ fa ⋅ ζ nfx −1
p
IE = o fa ⋅ ζ nfx
fx
(15)
(16)
where fx is a function of ex, and fa is a function of the geometric parameters. Table 1
summarizes the fx and fa functions for each of the cases.
Table 1 Indentation functions
Geometric Case
Case 1 :
Symmetric V
Wedge
Case 2 :
Symmetric Spoon
fx
fx = (3 + 2 ⋅ ex)
fa
fx = ((bex + 1)(1 + ex) + 1)
⎛ ⎛ 1 ⎞bex +1
⎞
c
⎟
⎟⎟
fa = ⎜ 2⎜⎜
⎜ ⎝ cos(γ ) ⎠
(bex + 1) sin(γ ) ⎟
⎝
⎠
Case 3 :
Right-Angle Edge
fx = (3 + 2 ⋅ ex)
⎞
⎛
1
⎟⎟
fa = ⎜⎜
2
⎝ sin(α ) cos(α ) sin( β `) cos ( β `) ⎠
Case 4 :
General Wedge
(Normal to hull)
Case 5 :
General Round
Edge
Case 6 :
Round
Vertical Cylinder
Case 7 :
Rectangular
Vertical Cylinder
Case 8 :
Spherical Contact
fx = (3 + 2 ⋅ ex)
⎛
⎞
tan(φ / 2)
⎟⎟
fa = ⎜⎜
2
⎝ sin( β `) cos ( β `) ⎠
fx = (2.5 + 1.5 ⋅ ex)
⎞
⎛
4
fa = ⎜⎜
2 R ⎟⎟
1.5
⎠
⎝ 3 ⋅ cos ( β `) sin( β `)
fx = (1.5 + 0.5 ⋅ ex)
fa = 2 H 2 R
fx = 1
fa = (HW )
fx = (2 + ex)
fa = (2 ⋅ π ⋅ R )
⎛
⎞
tan(α )
⎟⎟
fa = ⎜⎜
2
cos
(
γ
)
sin(
γ
)
⎝
⎠
1+ ex
1+ ex
(
1+ ex
1+ ex
1+ ex
)
1+ ex
1+ex
1+ ex
8. COLLISION TYPES
There are two types of collisions that can (presently) be solved by energy methods. The first is
a ‘normal’ type of impact. For general ship collisions this is referred to as a ‘Popov’ (see
Popov et. al. 1967) type of impact. The second is a beaching impact. Both will be described
and applied to various cases. Figure 9 shows a sketch of the two conditions, as they may exist
in a head-on ram. Either force may be larger, depending on the circumstances.
In the normal impact case the collision is idealized as a one-dimensional (normal) impact. The
normal kinetic energy is equated to the indentation energy. Potential energy (for example, due
to beaching) is ignored. Typically friction from sliding is also ignored. This type of analysis
can be used with any of the geometric cases described above, and for ship-ice and icestructure collisions. The analysis is valid within the range of the assumptions.
The beaching impact is a two-dimensional analysis. The total kinetic energy is equated to the
sum of indentation and potential energy. Again, friction is typically ignored. This type of
analysis can be used with geometric cases 1 and 2 described above, for ship-ice collisions.
And again, the analysis is valid within the range of the assumptions.
Figure 9. Collision conditions, initial impact and beached condition
9. INITIAL IMPACT COLLISIONS
A wide variety of collision scenarios can be analyzed as ‘normal’ collisions. The general
approach is presented followed by the force values that occur for the set of cases in Table 1.
Start by equating the normal kinetic energy with the ice crushing energy.
KEe = IE
(17)
where
KE e =
Me
⋅ Vn 2
2
(18)
which ,using equation (16) can be stated as;
KE e =
po
fa ⋅ ζ nfx
fx
(19)
Solving for the normal indentation:
1
⎛ KE ⋅ fx ⎞ fx
⎟⎟
ζ n = ⎜⎜ e
⋅
p
fa
⎝ o
⎠
(20)
The normal force can be found by substituting eqn. (20) into (15) to give
⎛ KE e ⋅ fx ⎞
⎟⎟
Fn = po ⋅ fa ⋅ ⎜⎜
p
fa
⋅
⎝ o
⎠
fx −1
fx
(21)
The values from Table 1 can be substituted into eqn. (21), together with (18) to get impact
force equations for each case. Table 2 shows the equations. The effective kinetic energy
depends on the nature of the collision. For head-on collisions (see Figure 10), the effective
kinetic energy (eqn. (18)) is the total kinetic energy. For oblique ship-ice collisions (see
Figure 11), the effective mass and velocity properties at the point of impact are determined as
follows;
Vn = Vship ⋅ lx
(22)
where Vn is the normal velocity at the point of impact
Vship is the x-direction velocity (all others are zero)
lx is the x-direction cosine
Me =
M ship
Co
where Me is effective mass at the point of impact
Mship is the ship’s mass (displacement)
Co is the mass reduction factor (see Popov et. al. 1967).
Co = l2/(1+AMx) + m2/(1+AMy) + n2/(1+AMz)
+ λ12/(rx2(1+AMrol) + µ12/(ry2 (1+AMpit)) + η12/(rz2 (1+AMyaw))
where;
l, m, n = direction cosines
λ1, µ1, η1 = roll, pitch, yaw, moment arms
AM~ = added mass factors
rx, ry, rz = mass radii of gyration
(23)
Table 2 Force equations – valid for initial impact (normal impact) conditions
Case 1:
1
Fn
p o.
(1
tan ( α )
ex )
. 1 . Me . Vn2 . ( 3
2
cos ( γ ) . sin ( γ )
2
2 .ex )
2 .ex )
(2
2 .ex )
(3
(3
2 . ex)
Case 2:
1
Fn
p o . 2..
( bex
1
1)
cos ( γ )
ex ) ( bex
(1
1 ) .( 1
ex )
1
c
.
. 1 . Me . Vn2 . ( ( bex
2
1 ) . sin ( γ ) )
( ( bex
( bex
1) .( 1
ex)
1) .
(1
( ( bex
1)
Case 3:
1
Fn
p o.
(1
1
sin ( α ) . cos ( α ) . sin ( β ' ) . cos ( β ' )
3
ex )
(2
2 .ex
. 1 . Me . Vn2 . ( 3
2
2
2 . ex)
(3
2 .ex )
2 .ex )
Case 4:
1
1.
tan
Fn
p o.
(1
φ
ex )
3
2 .ex
. 1 . Me . Vn2 . ( 3
2
2
sin ( β ' ) . cos ( β ' )
2
2 . ex)
(2
2 .ex )
(3
2 .ex )
Case 5:
1
Fn
p o.
4
3 . cos ( β ' )
1.5 .
. 2. R
(1
ex ) 2.5
( 1.5
1.5 .ex
. 1 . Me . Vn2 . ( 2.5 1.5. ex)
2
sin ( β ' )
Case 6:
1
Fn
p o . 2.H . 2 . R
(1
ex ) 1.5
.5 .ex
( .5
. 1 . Me . Vn2 . ( 1.5 .5. ex)
2
Case 7:
Fn
(1
p o .( H .W )
ex )
Case 8:
1
Fn
(1
p o . ( 2. π .R)
ex ) 2
ex
. 1 . Me . Vn2 . ( 2
2
ex)
(1
ex )
(2
ex )
( 1.5
.5 .ex )
.5 .ex )
( 2.5
1.5 .ex )
1.5 .ex )
ex )
1 ) .( 1
ex )
1)
Figure 10. Head-on (Symmetrical) Impact.
Figure 11. Shoulder (Oblique) Impact.
10. BEACHING IMPACT COLLISIONS
Head-on collisions between a ship and ice can result in the ship sliding up and beaching on the
ice. The general approach to these collisions is presented followed by the force values that
occur for the set of cases 1 and 2 in Table 1. To start we assume that initial kinetic energy is
equal to the sum of ice indentation (crushing) energy and pitch/heave potential;
KE = IE + PE
(24)
The kinetic energy is;
KE = 1/2 M V2
(25)
The potential energy, assuming linearity in heave and pitch is;
PE = 1/2 Fv2/Kb
(26)
where Kb is the effective vertical stiffness at the bow;
Kb = ρ g Awp /(1+(L/2/λ)2)
(27)
where
λ is the radius of gyration of the waterplane (i.e. Iwp = λ2 Awp).
Letting Kh = ρ g Awp, and assuming that, for most ships;
Kb = Kh /5
(28)
This gives;
5 Fv2
PE =
2 Kh
(29)
The verical force and normal force are related as:
(30)
Fv = Fn n
so that:
PE =
5 Fn2 ⋅ n 2
2 Kh
(31)
The force equation (15) can be re-written as:
Fn = K ice ⋅ ζ nfx −1
where
(32)
K ice = p o ⋅ fa
(33)
This allows the indentation energy equation to be written as:
IE =
K ice fx
ζn
fx
(34)
which with (32) can be rearranged to give:
−1
fx
(
K ice ) fx −1 fx −1
IE =
F
fx
(35)
n
The general beaching impact equation with (34) and (31) substituted into (24) is:
−1
1
5 Fn2 ⋅ n 2 (K ice ) fx −1 fx −1
M ⋅V 2 =
+
Fn
2
2 Kh
fx
fx
(36)
This equation can be solved for Fn . For certain special cases there is an analytical solution.
For the general cases, a numerical solution is required. For the simple linear case (for example
Case 1, ex= -.5) (36) reduces to:
Fn2
1
5 Fn2 ⋅ n 2
M ⋅V 2 =
+
2
2 Kh
2 ⋅ K ice
(37)
The force Fn can be solved for :
Fn =
1
5⋅n +
2
where:
κ=
K ice
1
⋅ M ⋅ Kh ⋅V
(38)
κ
K ice
Kk
⎛
⎞
tan(α )
⎟
= po ⋅ ⎜
⎜ cos 2 (γ ) sin(γ ) ⎟
⎝
⎠
0 .5
Kh = ρ g Awp
It is obvious that the beaching force, for the linear case, is proportional to velocity and the
square root of mass. It is also weakly dependant on the ice strength parameter po. Note that
this is just the solution for the beaching condition. The equivalent case for the initial impact
(with linear assumptions) is:
Fn = K ice ⋅
M
⋅V ⋅ l
Co
(39)
Comparison of (38) and (39) indicates that the beaching force is less dependent on hull form
than is the impact force.
11. DESIGN EQUATIONS
For longitudinal strength assessment equation (36) may be used as a simple check. It does not
include the effects of initial impact or any dynamics. Nevertheless, for cases which are
primarily beaching (i.e. large ships) the equation is valid. It is not analytically solvable for
most values of fx. A design equation could be formed by using equations for both impact and
beaching, covering a range of possible conditions. An equation of the form:
Fv = C1 ⋅ κ a ⋅ n b ⋅ M ⋅ K h ⋅ V
(40)
could be determined. Such an equation was first proposed by Riska (1994) (see also Daley and
Riska 1994). The constants C1, a, b would be determined by fitting the calculated results.
For oblique collisions equations from Table 2 may be used. A design equation based on Case
4 has been suggested for Polar Rule Harmonization work (see Kendrick and Daley 1998,
Daley 1999). The design equation for force has the form:
Fn = fa ⋅ po.36 ⋅ M .64 ⋅ V 1.28
(41)
where:
2
⎛
⎛x
⎞ ⎞ α
fa = ⎜ .097 − .68⎜ − .15 ⎟ ⎟
⎜
⎝L
⎠ ⎟⎠ β `
⎝
(42)
12. CONCLUSION
A variety of ice force equations have been presented. All are based on energy methods. These
results should be viewed as useful approximate values of force. More accurate results may be
obtained by solving the interaction equations directly, as a time series for instance.
Nevertheless, these energy solutions give insight into the process, particularly for cases in
which the energy balance governs the outcome.
The next important step required is the general solution of the oblique collision, with sliding
motions taken into account. While the impact idealization is essentially one-dimensional, and
the beaching collision is two-dimensional, the sliding-oblique collision is three-dimensional.
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